CODEWITHSUNDEEP
c++
Chris Bui
Chris Bui

Reputation: 1095

Vector Ranges in C++

Another quick question here, I have this code:

  string sa[6] = {
         "Fort Sumter", "Manassas", "Perryville",
         "Vicksburg", "Meridian", "Chancellorsville" };
  vector<string> svec(sa, sa+6);
  for (vector<string>::iterator iter = svec.begin(); iter != svec.end(); iter++)
  {
          std::cout << *iter << std::endl;
  }

Why is it that when I do svec(sa, sa+7), the code works but it prints out an empty line after the last word and when I do sa+8 instead it crashes? Because the string array is only 6 elements big, shouldn't it crash at sa+7 also?

Thanks.

Upvotes: 1

Views: 238

Answers (4)

Rob Kennedy
Rob Kennedy

Reputation: 163357

You have an array of only six elements. When you try to access the supposed "seventh" element, you get undefined behavior. Technically, that means anything can happen, but that doesn't seem to me like a very helpful explanation, so let's take a closer look.

That array occupies memory, and when you accessed the element beyond the end, you were reading whatever value happened to occupy that memory. It's possible that that address doesn't belong to your process, but it probably is, and so it's generally safe to read the sizeof(string) bytes that reside in that space.

Your program read from it and, since it was reading it through a string array, it treated that memory as though it were a real string object. (Your program can't tell the difference. It doesn't know any better. It's just trying to carry out your instructions.) Apparently, whatever data happened to be there looked enough like a real string object that your program was able to treat it like one, at least long enough to make a copy of it in the vector and then print its (empty) value. It worked this time, but that doesn't mean it will work every time.

There was no such luck with the data in the "eighth" position of the array. It did not look enough like a valid string object. A string object usually contains a pointer to the character data, along with a length. Maybe the area of the object that would normally represent that pointer didn't contain a valid address for your program. Or maybe the part that represented the length field contained a value far larger than what was available at the address in the pointer.

Upvotes: 5

ChrisW
ChrisW

Reputation: 56123

C++ doesn't do range-checking of arrays.

Reading beyond the end of an array is what's called "undefined" behaviour: i.e. it's not guaranteed to throw an exception, it's not guaranteed to not throw an exception, and it's not guaranteed to have consistent behaviour from one run to the next.

If people say that C++ is an "unsafe" language, this is part of what they mean by that. C++ doesn't check the range at run-time, because doing that a run-time take extra CPU instructions, and part of the design philosophy of C++ is to make it no slower than C.

Your compiler might have been able to warn you at compile-time (are you using the compiler command-line options to give you the maximum possible number of warnings?), though that too isn't guaranteed/required by the language.

Upvotes: 2

camh
camh

Reputation: 42538

Your application does not crash because there is some standard that specifies that it should crash. Crashing is just random (undefined) behaviour. You will not always get a crash when you exceed the bounds of an array as you have found out.

Essentially, anything could happen such as printing a blank line, crashing or even as just posted - have demons fly out of your nose.

Upvotes: 5

Fred Larson
Fred Larson

Reputation: 62113

Accessing past the end of a vector is undefined behavior. Anything could happen. You might have nasal demons.

Upvotes: 11

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